\(=\dfrac{1}{2} \begin{vmatrix} x _1&& x_2 &&x_3 & &x_1\\ &\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }&&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }\\ y_{1}&& y_2 & &y_3 && y_1 \end{vmatrix}\)
\(=\dfrac{1}{2}|(x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3)|\)
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Area of quadrilateral \(PQRS\)
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\(=\dfrac{1}{2} \begin{vmatrix} x _1&& x_2 &&x_3 & &x_4 & &x_1\\ &\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }&&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }\\ y_{1}&& y_2 & &y_3 && y_4 && y_1 \end{vmatrix}\)
\(=\dfrac{1}{2}\begin{vmatrix}(x_1y_2+x_2y_3+x_3y_4+x_4y_1)\\-(x_2y_1+x_3y_2+x_4y_3+x_1y_4) \end{vmatrix}\)
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| In general, when the coordinates of each vertex of a polygon are known, we can determine the area of the polygon using |
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\(=\dfrac{1}{2} \begin{vmatrix} x _1&& x_2 &&x_3 & &... & &x_n& &x_1\\ &\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }&&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } \\ y_{1}&& y_2 & &y_3 && ... && y_n && y_1 \end{vmatrix}\)
\(=\dfrac{1}{2}\begin{vmatrix}(x_1y_2+x_2y_3+...+x_ny_1)\\-(x_2y_1+x_3y_2+...+x_1y_n) \end{vmatrix}\)
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Example:
Find the area of the triangle with the vertices given.
\(A(-7,5),\, B(2,-4), \, C(4,3)\)
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| Based on the question, |
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\(A(-7,5),\, B(2,-4), \, C(4,3)\)
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Then, the area of the triangle \(\Delta{ABC}\)
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\(=\dfrac{1}{2} \begin{vmatrix} -7&& 2 &&4 & &-7\\ &\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }&&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }\\ 5&& -4 & &3 && 5 \end{vmatrix}\)
\(\begin{aligned} &=\dfrac{1}{2}\begin{vmatrix}(28+6+20)\\-(10-16-21)\end{vmatrix} \\\\ &=\dfrac{1}{2}|54+27| \\\\ &=40.5 \text{ unit}^2. \end{aligned}\)
